How does one show that the parametrization below maps exactly onto the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
$$\vec r(t)=\langle a\cos t, b \sin t \rangle$$
With $t \in [0,2\pi]$, it is easy to see that the equation of such ellipse is satisfied with such parametrization. Even more because $\frac{x^2}{a^2} \leq 1$ we see that it maps every $x$ value on the ellipse to a $y$ value satisfying the equation. But how do we know prove maps exactly the ellipse. __
I think I got an argument, for every value of $x$ there is at most two value of $y$, opposite in sign. If we let $a,b>0$ then for $t \in [0,\pi]$ we account for all values of $x$ but by our parametrization $y$ is nonnegative. Then for $t \in [\pi,2\pi]$ we again account for all values of $x$ but this time $y$ is nonpositive. This seems convincing to me, I don't know how to formally write this or explain why this proves what I'm trying to say.
Let $x,y\in\mathbb R$ be such that $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Define $X=\frac xa$ and $Y=\frac yb$. Then $X^2+Y^2=1$ and so $X=\cos\theta$ and $Y=\sin\theta$ for some $\theta\in[0,2\pi]$. But then $x=aX=a\cos\theta$ and $y=bY=b\sin\theta$.
If you do not see why is it that $X^2+Y^2=1$ implies that $X=\cos\theta$ and $Y=\sin\theta$ for some $\theta\in[0,2\pi]$, here's a proof. First of all, $X^2+Y^2=1\Longrightarrow X^2\leqslant1\Longleftrightarrow-1\leqslant X\leqslant1$. So, by the intermediate value theorem, $X=\cos\theta'$ for some $\theta'\in[0,\pi]$. If it turns out that $\sin\theta'=Y$, then we're done: $\theta=\theta'$. Otherwise, since $\sin^2\theta'=1-\cos^2\theta'=1-X^2=Y^2$, then $\sin\theta'=-Y$. So, take $\theta=2\pi-\theta'$. Then $\cos\theta=\cos\theta'=X$ and $\sin\theta=-\sin\theta'=Y$. Furthermore, $\theta\in[\pi,2\pi]\subset[0,2\pi]$-